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horrorfan [7]
2 years ago
14

The equation Upper E equals StartFraction 1 Over 2 pi epsilon Subscript 0 EndFraction StartFraction q ⁢ d Over z Superscript 3 E

ndFraction is approximation of the magnitude of the electric field of an electric dipole, at points along the dipole axis. Consider a point P on that axis at distance z = 23.00d from the dipole center (d is the separation distance between the particles of the dipole). Let Eappr be the magnitude of the field at point P as approximated by the equations below. Let Eact be the actual magnitude. What is the ratio Eappr /Eact?
Physics
1 answer:
Allushta [10]2 years ago
4 0

Answer:

 E_aprox = 1.003 E_real

Explanation:

In this exercise we are given the expression for the electric field of a dipole in the axis direction of the dipole

         E_real = k 2q d / √(z² + d²)³

         

I think your equation has some errors.

In this case they indicate that d is the separation of the charges of the dipole

in the case of z »d this equations approximates

        E_aprox = k 2q  d/ z³

calculate the value for the two cases

         E_real = k2q  d / √[ ((23d)² + d²)³]  

         E_real = k2q  d / d³ 12201

         E_real = k2q   1/12201 d²

         E_aprox = k2q   d / (23.00d)³

         E_aprox = k2q   1/12167 d²

         

    the error between these quantities is

        E_aprox / E_real = 12201 d² / 12167 d²

        E_aprox / E_real = 1.003

        E_aprox = 1.003 E_real

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This question is incomplete, the complete question is;

Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.  

1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?

Fwith belt =

2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.

Fwithout belt =

Answer:

1) The Net force on the driver with seat belt is 10.3 KN

2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

Given the data in the question;

from the equation of motion, v² = u² + 2as

we solve for a

a = (v² - u²)/2s ----- let this be equation 1

we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

1)

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

so we substitute the given values into the equation;

F = 70( ((0)² - (18)²) / 2 × 1.1 )

F = 70 × ( -324 / 2.4 )

F = 70 × -147.2727

F = -10309.09 N

F = -10.3 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwith belt =  10.3 KN

Therefore, Net force of the driver is 10.3 KN

2)

No sit belt,  

m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m

we substitute

F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )

F = 70 × ( -324 / 0.022 )

F = 70 × -14727.2727

F = -1030909.08 N

F = -1030.9 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwithout belt = 1030.9 KN

Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

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